Question: $f(x, y) = \dfrac{x^2}{2} - x^4 + \cos(x) - y^2 + 2y$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-0.511 \dots, 1)$ (Choice B) B $(0, 1)$ (Choice C) C $(0.511 \dots, 1)$ (Choice D) D There are no critical points.
Answer: A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} x - 4x^3 - \sin(x) \\ \\ -2y + 2 \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} x - 4x^3 - \sin(x) = 0 \\ \\ -2y + 2 = 0 \end{cases}$ The equation $-2y + 2 = 0$ implies that $y = 1$. The second equation is $x - 4x^3 = \sin(x)$. One solution is $x = 0$. Are there any others? There are no other solutions for $x$. We know this because the left hand side has a derivative of $1$ at $0$, just like $\sin(x)$, but its third order derivative is far too high. The graph of $x - 4x^3$ moves away from $\sin(x)$ rapidly, and they never intersect aside from $x = 0$. ${0.5}$ ${1}$ ${1.5}$ ${\llap{-}0.5}$ ${\llap{-}1}$ ${\llap{-}1.5}$ ${0.5}$ ${1}$ ${1.5}$ ${\llap{-}0.5}$ ${\llap{-}1}$ ${\llap{-}1.5}$ $y$ $x$ Therefore, $f$ has a critical point at $(0, 1)$.